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\begin{fcthm}[]
  \label{thm_2_5}
  % Theorem 2 5
  Assuming:
  - $0 < \eps < \quarter$
  - $A \subset \Q_n$
  - $\frac{|A|}{2^n} \ge \half$
  Then:
  $\frac{|A_{\ineigh{\eps n}}|}{2^n} \ge 1 -
  \frac{2}{\eps} e^{- \frac{\eps^2 n}{2}}$.
\end{fcthm}

``$\half$-sized sets have exponentially large
$\eps n$-\glspl{neigh}.''

\begin{proof}
  Enough to show that if $\eps n$ an integer then
  \[
    \frac{|A_{\ineigh{\eps n}}|}{2^n} \ge 1 -
    \frac{1}{\eps} e^{- \frac{\eps^2 n}{2}}
  .\]
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/aaa592e8d5da46b2.png}
  \end{center}
  Have $|A| \ge \sum_{i = 0}^{\left\lceil
  \frac{n}{2} - 1 \right\rceil} {n \choose i}$, so
  by \nameref{harper}, we have $|A_{\ineigh{\eps
  n}}|
  \ge \sum_{i = 0}^{\left\lceil \frac{n}{2} - 1 +
  \eps n\right\rceil} {n \choose i}$, so
  \[
    |A_{\ineigh{\eps n}}^c| \le \sum_{i = \left\lceil
    \frac{n}{2} + \eps n \right\rceil}^n {n \choose
    i} = \sum_{i = 0}^{\left\lfloor \frac{n}{2} - \eps
    n\right\rfloor} {n \choose i} \le \frac{1}{\eps}
    e^{- \frac{\eps^2 n}{2}} 2^n
    . \qedhere
  \]
\end{proof}

\begin{remark*}
  Same would show, for ``small'' sets:
  \[
    \frac{|A|}{2^n} \ge \frac{2}{\eps} e^{-
    \frac{\eps^2 n}{2}}
    \implies \frac{|A_{\ineigh{2\eps n}}|}{2^n}
    \ge 1 - \frac{2}{\eps} e^{- \frac{\eps^2 n}{2}}
  .\]
\end{remark*}

\subsection{Concentration of measure}

\glsadjdefn{lips}{Lipschitz}{function}%
Say $f : \Q_n \to \Rbb$ is \emph{Lipschitz} if
$|f(x) - f(y)| \le 1$ for all $x, y$ adjacent.

\glsadjdefn{med}{median}{function}%
For $f : \Q_n \to \Rbb$, say $M \in \Rbb$ is a
\emph{L\'evy mean} or \emph{median} of $f$ if
\[
  |\{x \in \Q_n : f(x) \le M\}| \ge 2^{n - 1}
\]
and
\[
  |\{x \in \Q_n : f(x) \ge M\}| \ge 2^{n - 1}
\]

Now ready to show ``every well-behaved function on
the cube $\Q_n$ is roughly constant nearly
everywhere''.

\begin{fcthm}[]
  % Theorem 2 6
  Assuming:
  - $f : \Q_n \to \Rbb$ \gls{lips} with \gls{med}
  $M$
  Then:
  \[
    \frac{|\{x : |f(x) - M| \le \eps n\}|}{2^n}
    \ge 1 - \frac{4}{\eps} e^{- \frac{\eps^2 n}{2}}
  \]
  for any $0 < \eps < \frac{1}{4}$.
\end{fcthm}

\begin{note*}
  This is the ``concentration of measure''
  phenomenon.
\end{note*}

\begin{center}
  \includegraphics[width=0.3\linewidth]{images/e569b56ea85b4e22.png}
\end{center}

\begin{proof}
  Let $A = \{x : f(x) \le M\}$. Then
  $\frac{|A|}{2^n} \ge \half$, so
  \[
    \frac{|A_{\ineigh{\eps n}}|}{2^n} \ge 1 -
    \frac{2}{\eps} e^{- \frac{\eps^2 n}{2}}
  .\]
  ut $f$ is \gls{lips}, so $x \in A_{\ineigh{\eps
  n}}$ implies $f(x) \le M + \eps n$. Thus
  \[
    \frac{|\{x : f(x) \le M + \eps n\}|}{2^n} \ge
    1 - \frac{2}{\eps} e^{- \frac{\eps^2 n}{2}}
  .\]
  Similarly,
  \[
    \frac{|\{x : f(x) \ge M - \eps n\}|}{2^n} \ge
    1 - \frac{2}{\eps} e^{- \frac{\eps^2 n}{2}}
  .\]
  Hence
  \[
    \frac{|\{x : M - \eps n \le f(x) \le M + \eps
    n\}|}{2^n} \ge 1 - \frac{4}{\eps} e^{-
    \frac{\eps^2 n}{2}}
    . \qedhere
  \]
\end{proof}

Let $G$ be a graph of diameter $D$ ($D = \max \{f(x,
y) : x, y \in G\}$).

\begin{fcdefnstar}[$alpha(G, eps)$]
  \glssymboldefn{alphaG}%
  Write
  \[
    \alpha(G, \eps) = \max \left\{ 1 -
    \frac{|A_{\ineigh{\eps D}}|}{|G|} : A \subseteq G,
    \frac{|A|}{|G|} \ge \half \right\}
  .\]
  So $\alpha(G, \eps)$ small says ``$\half$-sized
  sets have large $\eps D$-neighbourhoods''.
\end{fcdefnstar}

\begin{fcdefnstar}[Levy family]
  % [L\'evy family]
  \glsnoundefn{lf}{L\'evy family}{L\'evy families}%
  Say a sequence of graphs is a \emph{L\'evy family}
  if $\alphaG(G_n, \eps) \to 0$ as $n \to \infty$,
  for each $\eps > 0$.
\end{fcdefnstar}

So \cref{thm_2_5} tells us that the sequence
$(\Q_n)$ is a \gls{lf} -- even a \emph{normal
L\'evy family}, meaning $\alphaG(G_n, \eps)$ grows
exponentially small in $n$, for each $\eps > 0$.

So have concentration of measure for any \gls{lf}.

Many naturally-occurring families of graphs are
\glspl{lf}.

\begin{example*}
  $(S_n)$, where $S_n$ is made into a graph by
  joining $\sigma$ to $\sigma'$ if $\sigma'
  \sigma^{-1}$ is a transposition.
\end{example*}

Can define $\alphaG(X, \eps)$ similarly for any
metric measure space $X$ (of finite measure and
finite diameter).

\begin{example*}
  $(S^n)$ is a \gls{lf}.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/4409a430bb404d2e.png}
  \end{center}
  Two ingredients:
  \begin{enumerate}[(1)]
    \item An isoperimetric inequality on $S_n$:
      for $A \subset S_n$, $C$ a circular cap with
      $|C| = |A|$, have $|A\neigh{\eps}| \ge
      |C_{\ineigh{\eps}}|$.

      Proof by compression:
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/c89c8d437e7444d3.png}
      \end{center}
    \item Estimate: circular cap $C$ of measure
      $\half$ is the cap of angle $\frac{\pi}{2}$,
      so $C_\eps$ is the circular cap of angle
      $\frac{\pi}{2} + \eps$.
      \begin{center}
        \includegraphics[width=0.3\linewidth]{images/de9624c0e3574a84.png}
      \end{center}
      This complement has measure about
      $\int_{\eps}^{\frac{\pi}{2}} \cos^{n - 1} t \dd t$, which
      $\to 0$ as $n \to \infty$.
      \begin{center}
        \includegraphics[width=0.3\linewidth]{images/7a02fd0102984c3e.png}
      \end{center}
  \end{enumerate}
\end{example*}

We deduced concentration of measure from an
isoperimetric inequality.

Conversely:

\begin{fcprop}[]
  % Proposition 2 7
  Assuming:
  - $t$, $\alpha$ and $G$ are such that for any \gls{lips}
  function $f : G \to \Rbb$ with \gls{med} $M$ we
  have
  \[
    \frac{|\{x \in G : |f(x) - M| > t\}|}{|G|} \le
    \alpha
  ,\]
  Then:
  for all $A \subseteq G$ with $\frac{|A|}{|G|}
  \ge \half$, we have $\frac{|A_{\ineigh{t}}|}{|G|}
  \ge 1 - \alpha$.
\end{fcprop}

\begin{proof}
  The function $f(x) = d(x, A)$ is \gls{lips}, and
  has $0$ as a \gls{med}, so
  \[
    \frac{|\{x \in G : x \notin
    A_{\ineigh{t}}\}|}{|G|} \le \alpha
    . \qedhere
  \]
\end{proof}